Question: Simplify and expand the following expression: $ \dfrac{2}{2z - 4}+ \dfrac{1}{z + 8}+ \dfrac{3}{z^2 + 6z - 16} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{2}{2z - 4} = \dfrac{2}{2(z - 2)}$ We can factor the quadratic in the third term: $ \dfrac{3}{z^2 + 6z - 16} = \dfrac{3}{(z - 2)(z + 8)}$ Now we have: $ \dfrac{2}{2(z - 2)}+ \dfrac{1}{z + 8}+ \dfrac{3}{(z - 2)(z + 8)} $ The least common multiple of the denominators is: $ 2(z - 2)(z + 8)$ In order to get the first term over $2(z - 2)(z + 8)$ , multiply by $\dfrac{z + 8}{z + 8}$ $ \dfrac{2}{2(z - 2)} \times \dfrac{z + 8}{z + 8} = \dfrac{2(z + 8)}{2(z - 2)(z + 8)} $ In order to get the second term over $2(z - 2)(z + 8)$ , multiply by $\dfrac{2(z - 2)}{2(z - 2)}$ $ \dfrac{1}{z + 8} \times \dfrac{2(z - 2)}{2(z - 2)} = \dfrac{2(z - 2)}{2(z - 2)(z + 8)} $ In order to get the third term over $2(z - 2)(z + 8)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{3}{(z - 2)(z + 8)} \times \dfrac{2}{2} = \dfrac{6}{2(z - 2)(z + 8)} $ Now we have: $ \dfrac{2(z + 8)}{2(z - 2)(z + 8)} + \dfrac{2(z - 2)}{2(z - 2)(z + 8)} + \dfrac{6}{2(z - 2)(z + 8)} $ $ = \dfrac{ 2(z + 8) + 2(z - 2) + 6} {2(z - 2)(z + 8)} $ Expand: $ = \dfrac{2z + 16 + 2z - 4 + 6}{2z^2 + 12z - 32} $ $ = \dfrac{4z + 18}{2z^2 + 12z - 32}$ Simplify: $ = \dfrac{2z + 9}{z^2 + 6z - 16}$